Ok, I posted the following question: Consider this reaction: 4 KO2 (s) + 2 CO2 (

Question

Ok, I posted the following question:
Consider this reaction: 4 KO2 (s) + 2 CO2 (g) ? 2 K2CO3(s) + 3 O2 (g)
(a) How many moles of oxygen gas are produced when 0.400 mol of KO2 reacts with excess carbon dioxide?
(b) How many grams of KO2 are needed to form 7.50 g of oxygen gas?
(c) How many grams of CO2 are used when 7.50 g of oxygen are produced?

And got the following answers:
a.0.3mole
b.(7.5/32)x(4/3)=0.3125mole =22.185gm
c.(7.5/32)x(2/3)=0.15626mole =6.875gm

Now I understand the 7.5 is the 7.50 g of oxygen and the 32 is the mass of O2. The 4 is KO2 and the 3 the O2. However, I don't understand how he turned the moles to grams

Explanation / Answer

b) molar mass of KO2 = 71.1 g per mole for 0.3125 moles = 0.3125 x 71.1 = 22.22 g c) molar mass of CO2 = 44 0.15626 moles = 0.15626 x 44 = 6.875 g

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